Question

oppositlely charged parallel plates are separated by 5.00 mm. A potential difference of 400.0 v exists between the plates. a) what is the magnitude of the electric field between the plates? b) how much work does the electric field do moving an electron from the negative plate to the positive plate?

Answer 1

distance between the plates d = 5mm = 0.005 m

potential difference V = 400 V

a) magnitude of electric field between the plates, E

E = 80,000 V/m = 80,000 N/C

b) work done by the electric field to move an electron from negative plate to positive plate is:

W = Fd;

where d is the distance between the plates;

F is the force on an electron due to the electric field and is given by F=qE, q is the charge on an electron;

F = (1.602 * 10^-19 C) (80,000 N/C)

   = 1.28*10^-14 N

W = (1.28*10^-14 N)(0.005m)

     = 6.41 * 10^-17 J